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Unformatted text preview: homework 02 – HOFFMA, DAVID – Due: Jan 28 2008, 4:00 am 1 Question 1, chap 23, sect 2. part 1 of 3 10 points A uniformly charged circular arc AB of ra dius R is shown in the figure. It covers a quarter of a circle and it is located in the sec ond quadrant. The total charge on the arc is Q > 0. The value of the Coloumb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . x y + + + + + + + + + Δ θ θ R x y I II III IV B A O Δ s ≡ R Δ θ What is the direction of the electric field vector vector E at the origin, due to the charge distribution? 1. in quadrant III 2. along the positive xaxis 3. along the negative yaxis 4. in quadrant IV correct 5. along the negative xaxis 6. along the positive yaxis 7. in quadrant I 8. in quadrant II Explanation: The electric field for a positive charge is directed away from it. In this case, the electric field generated by each Δ q will be directed into quadrant IV, so the total electric field will be in the same quadrant. Question 2, chap 23, sect 2. part 2 of 3 10 points Find E x , the xcomponent of the electric field at the origin due to the full arc length for the case, where Q = 2 . 5 μ C and R = 1 . 42 m. Correct answer: 7093 . 9 N / C (tolerance ± 1 %). Explanation: Let : Q = 2 . 5 μ C = 2 . 5 × 10 − 6 C , R = 1 . 42 m , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . E x can be found by integrating the contri butions of all the Δ q ’s in the arc. Using Δ q = Δ θ parenleftBig π 2 parenrightBig and taking the xcomponent, then adding the force component from each charge Δ q at angle θ , we have E x = integraldisplay π/ 2 k e Q R 2 2 π cos θ dθ = 2 k e Q π R 2 = 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) π (1 . 42 m) 2 × (2 . 5 × 10 − 6 C) = 7093 . 9 N / C . Question 3, chap 23, sect 2. part 3 of 3 10 points If at O we have E x = C , then what is the magnitude of the full force F (not just the xcomponent) on an electron at this point? 1. F = C 3 e 2. F = 2 C 3 e 3. F = 3 e C 2 4. F = C e √ 2 5. F = 2 e C 6. F = C 2 e homework 02 – HOFFMA, DAVID – Due: Jan 28 2008, 4:00 am 2 7. F = e C 8. F = 3 e C 9. F = 4 e C 10. F = √ 2 e C correct Explanation: From the symmetry of the charge distri bution, it can be seen that E y = E x = C , so E x E y E E = radicalBig E 2 x + E 2 y = √ 2 C . The force on an electron is then F = e E = √ 2 e C . Question 4, chap 23, sect 2. part 1 of 4 10 points Consider a disk of radius 2 . 5 cm, having a uniformly distributed charge of +4 . 2 μ C. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Compute the magnitude of the electric field at a point on the axis and 3 . 6 mm from the center. Correct answer: 1 . 03576 × 10 8 N / C (tolerance ± 1 %)....
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This note was uploaded on 03/19/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Charge, Work

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